I'm having trouble with this double integral:

\[\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x)\sin (y)\sin(x+y)}{xy(x+y)}\mathrm {d}x \mathrm {d}y\]

The result of the mathematica seems to be $\frac{\pi^2}{6}$.But I have no idea how to solve it.

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## How to calculate the double integral？

I'm having trouble with this double integral:

\[\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x)\sin (y)\sin(x+y)}{xy(x+y)}\mathrm {d}x \mathrm {d}y\]

The result of the mathematica seems to be $\frac{\pi^2}{6}$.But I have no idea how to solve it.

Write the integral as

$$

I = \mathrm{Im} \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{i(x+y)}

$$

Define

$$

J(a) \equiv \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{ia(x+y)}

$$

so

$$

J'(a) = i \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy}e^{ia(x+y)} \equiv i (K(a))^2,

$$

where

$$

K(a) = \int_0^\infty dx \frac{\sin x}{x}e^{i a x}

$$

We have, for $\mathrm{Im}\ a = \epsilon > 0$,

$$

K'(a) = i\ \int_0^\infty dx \sin x\ e^{i a x} = \frac i 2 \left[\frac{1}{a+1} - \frac{1}{a-1} \right]

$$

so

$$

K(a) = \int_{\infty + \epsilon i}^{a} dz\ K'(z) = i\ \mathrm{arctanh}(1/a)

$$

We have, for example from Riemann-Lebesgue,

$$

J(\infty+\epsilon i) = 0

$$

whence

$$

J(1) = \int_\infty^1 da\ J'(a)

$$

Taking the imaginary part, we get

$$

I = \int_1^\infty dx\ \mathrm{arctanh}^2 \frac 1 x = \int_0^1 \frac{dx}{x^2}\ \mathrm{arctanh}^2 x \\= \frac 1 4 \int_0^1 dx\ \frac{\ln^2(1+x)+\ln^2(1-x)-2 \ln(1+x)\ln(1-x)}{x^2}

$$

The integral can be evaluated by finding the antiderivative expressed in terms of the polylogarithm, or by a series expansion. I don't have time to write down all the steps, which are standard. The result is

$$

I = \frac{\pi^2}{6}

$$

**Moderators:** galactus, Random Variable, sos440

2 posts
• Page **1** of **1**

\[\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x)\sin (y)\sin(x+y)}{xy(x+y)}\mathrm {d}x \mathrm {d}y\]

The result of the mathematica seems to be $\frac{\pi^2}{6}$.But I have no idea how to solve it.

$$

I = \mathrm{Im} \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{i(x+y)}

$$

Define

$$

J(a) \equiv \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{ia(x+y)}

$$

so

$$

J'(a) = i \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy}e^{ia(x+y)} \equiv i (K(a))^2,

$$

where

$$

K(a) = \int_0^\infty dx \frac{\sin x}{x}e^{i a x}

$$

We have, for $\mathrm{Im}\ a = \epsilon > 0$,

$$

K'(a) = i\ \int_0^\infty dx \sin x\ e^{i a x} = \frac i 2 \left[\frac{1}{a+1} - \frac{1}{a-1} \right]

$$

so

$$

K(a) = \int_{\infty + \epsilon i}^{a} dz\ K'(z) = i\ \mathrm{arctanh}(1/a)

$$

We have, for example from Riemann-Lebesgue,

$$

J(\infty+\epsilon i) = 0

$$

whence

$$

J(1) = \int_\infty^1 da\ J'(a)

$$

Taking the imaginary part, we get

$$

I = \int_1^\infty dx\ \mathrm{arctanh}^2 \frac 1 x = \int_0^1 \frac{dx}{x^2}\ \mathrm{arctanh}^2 x \\= \frac 1 4 \int_0^1 dx\ \frac{\ln^2(1+x)+\ln^2(1-x)-2 \ln(1+x)\ln(1-x)}{x^2}

$$

The integral can be evaluated by finding the antiderivative expressed in terms of the polylogarithm, or by a series expansion. I don't have time to write down all the steps, which are standard. The result is

$$

I = \frac{\pi^2}{6}

$$

2 posts
• Page **1** of **1**